高等化工热力学-第一章

热机，等做的功。因此
W Ws [( PV )m ]
d (mU ) dt
Q Ws [(U 1 u2 gz )m ] [( PV )m ] 2
Q Ws [(U PV 1 u2 gz )m] 2
因为 H = U + PV，得到
Note: Hin is constant and the subscript i denotes the initial state.
Integration
H in U i n H in U ni
Considering the air as an ideal gas, then n PTi ni PiT
d (mU ) dt
Q Ws [( H 1 u2 gz )m ] 2
d ( mU ) dt
[( H
1 u2 2
gz )m ] Q Ws
这就是敞开体系热力学第一定律的通用形式。
Problem
An insulated, electrically heated tank for hot water contains 190 kg of liquid water at 60 0C when a power outage occurs. If water is withdrawn from the tank at a steady rate of 0.2 kg/s, how long will it take for the temperature of the water in the tank to drop from 60 to 35 0C? Assume that cold water enters the tank at 10 0C, and Cv=Cp=C, independent of T and P. (一个保温很好、容量为190kg 的电热水器将水加热到60 0C 时，突然断电。如果此时电热水器以0.2 kg/s的质量流速放 出热水，问需要多长时间电热水器里面的水由60 0C 降到 35 0C？假设：电热水器进口的冷水温度为10 0C，水的CV = CP = C，与温度无关。)
Since
dU CV dT CdT dH C P dT CdT H H 1 C (T T1 )
Therefore
m dT dt m T T1
m T T1 ln( ) m T0 T1
Integrating the above equation from t = 0 to arbitrary time t yields
Solution
1st Step: 确定体系 It is obvious that we should consider the water in the tank as the system. Therefore the system in this problem is an open system.
Chapter 1 Elements of Thermodynamics 热力学基础
Part Ⅰ 热力学第一定律和第二定律
The First Law and Second Law of Thermodynamics
体系的分类：
封闭体系(closed system)：与环境没有物质交换。 体系 敞开体系(open system)：与环境有质量交换。 简单体系（simple system)：体系内部没有任何绝热 (adiabatic)、刚性(rigid)、不渗透(impermeable)的界 面(boundary)，而且不受外部和内部力场的作用。 复杂体系(composite system)：体系由两个或两个以 上的简单体系构成，如半渗透膜海水淡化体系；或 者体系处于外部或内部力场中，如磁场、电场等。 简单体系是最常见的一种体系。对于简单体系，体系的 能量等于其内能，即 E = U。
Figure The schematic of open system
对于给定的敞开体系 进入体系 的能量
1 2 流入体系的物质带入的能量，(U u gz )1 m1 2
体系与环境交换的热和功，Q W
1 2 离开体系的能量: 离开体系的物质帯走的能量, (U u gz )2 m2 2 体系能量的积累 = 进入体系的能量-离开体系的能量 d ( mU ) 1 1 (U u 2 gz )1 m1 (U u 2 gz )2 m2 Q W dt 2 2 W [(U 1 u 2 gz )m ] Q 2
体系
Section 1 The First Law of Thermodynamics
热力学第一定律是对自然界普遍存在的能量守衡(energy conversation)规律的总结，可以用下面的数学公式表示： △(Energy of the system) +△(Energy of surroundings) = 0 物理化学告诉我们封闭体系的热力学第一定律是
Note: Δ = 2-1(Exit - Entrance)
其中， 是流体的质量流速，kg/s；Q 是热交换速率，J/s；W m
是功率。对于界面是刚性的体系，不存在体积功， 只包括两 W
( PV 部分，一是流体进入和离开体系所做的流动功率 )m
；另
一种是轴功率(shaft work rate)， ，它是通过机械装置，如泵， Ws
t
Substitution of numerical values into this equation, gives
190 35 10 t ln( ) 658.5 s 0.2 60 10
在另外一些情况下，如果体系的部分界面可以移动，则 体系的体积要发生变化，体系与环境之间有体积功的交换。 这时，如果忽略体系的动能差和势能差，敞开体系的热力学 第一定律可以写成下面的形式
d ( mU ) dt
[( H 1 u2 gz )m ] Q Ws 2
3rd Step: 分析热力学第一定律中的各个项 In this problem, the heat rate is zero because a power outage occurs and heat losses from the tank are negligible. The work rate is also zero since there is no mechanical equipment in the process. The mass in the tank and the mass flowrate are constant. Moreover, if the differences between inlet and outlet kinetic and potential energies can be neglected, the equation of 1st law is simplified as the following form: dU m m( H H1 ) 0 dt where U and H refer to the internal energy and enthalpy of the water in the tank; H1 is the enthalpy of the cold water entering the tank.
Solution
Step (a): The definition of system:The most convenient system is the gas in the tank at any time. This is then an open system at constant total volume. d ( nU ) H in dnin H out dnout Q Ws W
2nd Step: 应用敞开体系热力学第一定律 Since the system is an open system, we should use the equation of the general first law of thermodynamics for open systems, to solve the problem
(nU ) Q W
而在实际化工过程或物理过程中遇到的体系不一定都是封闭 体系，通常是敞开体系。敞开体系的热力学第一定律如何表 达？ 下面是敞开体系的示意图。
Q
2(out) m 2 u2 P2, T2
m1 1(in) u1
H1, U1
z1 水平线
z2
d (nU )
in
H in dnin


out
H out dnout Q Ws W
Problem
Process: A 4 m3 storage tank containing 2 m3 of liquid is about to be pressurized with air from a large, high-pressure reservoir through a valve at the top of the tank to permit rapid ejection of the liquid (see the following figure) . The air in the reservoir is maintained at 100 bar and 300 K. The gas space above the liquid contains air at 1 bar and 280 K. When the pressure in the tank reaches 5 bar, the liquid transfer valve is opened and the liquid is ejected at the rate of 0.2 m3/min while the tank pressure is maintained at 5 bar.
Storage Tank Air Reservoir
Liquid Transfer Valve
Questions: What is the air temperature in the storage tank when the pressure reaches 5 bar and when the liquid has been drained completely?
H in U i PTi H in U PiT
For ideal gas,
dH CPdT
dU C V dT
The subscript 0 denotes the reference state.
H in C P (Tin T0 ) H 0 U i C V (Ti T0 ) U 0
U C V (T T0 ) U 0
H in U i C PTin C VTi (C P C V )T0 H 0 U 0
H in U C PTin C VT (C P C V )T0 H 0 U 0
in out
Since
Why?
d (nU ) Udn ndU
Q Ws W 0
Reservoir lager relative to system
Hin constant
dnout 0
dnin dn
Then
Udn ndU Hindn
dU dn H in U n